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Unit 12: Fundamental Theorem of Algebra




          Therefore,                                                                            Notes

                                g(z)   =  (|a | |a | ... |a n 1 |)r n 1
                                                          
                                                
                                                  
                                           
                                         0
                                              1
                                                      
                                f(z)          |a |r n
                                                n
                                      |a | |a | ... |a  |
                                                 
                                               
                                          
                                     =  0    1      n 1
                                                     
                                            |a |r
                                              n
          Hence |g(z) < |f(z)|, provided that
                   |a | |a | ... |a n 1 |  < 1
                       
                            
                              
                          1
                     0
                                  
                         |a |r
                           n
                                      |a | |a | ... |a  |
                                          
                                                 
                                               
          i.e.                     r >  0    1      n 1                             (1)
                                                     
                                             |a |
                                               n
          Since r is arbitrary, therefore, we can choose r large enough so that (1) is satisfied. Now, applying
          Rouche’s theorem, we find that the given polynomial f(z + g(z) has the same number of zeros as
          f(z). But f(z) has exactly n zeros all located at z = 0. Hence, the given polynomial has exactly n
          zeros.
                 Example: Determine the number of roots of the equation
                                 z  – 4z  + z  – 1 = 0
                                      5
                                  8
                                         2
          that lie inside the circle |z| = 1
          Solution. Let C be the circle defined by |z| = 1
          Let us take f(z) = z  – 4z , g(z) = z – 1.
                             5
                         8
                                     2
          On the circle C,
                       g(z)    z  1    |z| 1
                                          2
                                2
                                            
                       f(z)   =   z  4z 5   |z||4 z |
                                              3
                               8
                                         5
                                           
                              1 1     2   2
                               
                                          1
                             4 |z| 3  4 1  3
                                      
                              
          Thus, |g(z)| <|f(z)| and both f(z) and g(z) are analytic within and on C, Rouche’s theorem
          implies that the required number of roots is the same as the number of roots of the equation
          z  – 4z  = 0 in the region |z| < 1. Since z  – 4  0 for |z| < 1, therefore, the required number of
           8
               5
                                           3
          roots is found to be 5.
          Inverse Function
          If f(z) = w has a solution z = F(w), then we may write
          f{F(w) } = w, F{ f(z)} = z. The function F defined in this way, is called inverse function of f.
          Theorem. (Inverse Function Theorem)
          Let a function w = f(z) be analytic at a point z = z  where f (z )  0 and w  = f(z ).
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                                                 0
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